Hello pohoatza, pohoatza wrote: Does there exist a function $f: \mathbb{R}\rightarrow\mathbb{R}$ such that $f(x+f(y))=f(x)+\sin y$, for all reals $x,y$ ? 1) f(x) is surjective : 1.1) : f(x+f(y)) = f(x) + sin(y) ==> f(f(x)) = f(0) + sin(x) ==> f(R) contains [f(0)-1,f(0)+1] 1.2) : f(x+f(y)) = f(x) + sin(y) ==> f(x+nf(y)) = f(x) + n*sin(y) ==> f(x+nf(pi/2)) = f(x)+n and f(x+nf(-pi/2)) = f(x)-n So f(R) contains one interval of width 2 (see 1.1) and any translation by +/- n of such an interval (see 1.2) ==> f(R) = R Q.E.D. 2) f is bounded f is surjective ==> it exists x0 such that f(x0) = 0 f is surjective ==> for any x, it exists z such that f(z) = x-x0 Then : f(x) = f(x0 + f(z)) = f(x0) + sin(z) = sin(z) ==> f(x) is in [-1,+1] for any x Q.E.D. 3) f doesn't exist Obvious since 1) and 2) are in contradiction. -- Patrick

substitute $ 0$ instead of $ x$ into the main equation: $ f(f(x)) = f(0) + \sin{x}$ so $ f(f(x))\in [f(0) - 1,f(0) + 1]$ It means that a boundary of all possible values of $ f(f(x))$ is limited, consequently, the same claim should hold for $ f(x)$. so if there is some $ x$ such that $ f(x) = y$, then: $ f(x + f(z)) = y + \sin{z}$, so for every $ y_0\in[y - 1,y + 1]$ there should exist $ x_0$ such that $ f(x_0) = y_0$, it is enough to take $ x_0 = x + f(z_0)$, where $ \sin{z_0} = y_0 - y$. So the image of $ f$ can be infinitely extended, thus the boundary of its values is unlimited. Contradiction.

Erken wrote: It means that a boundary of all possible values of $ f(f(x))$ is limited, consequently, the same claim should hold for $ f(x)$. I think this is not right... is it? If this statement is true, why?

Clearly wrong : Let $f(x)=|x|-x$ so that $f(f(x))\equiv 0$ We have there an example of unbounded continuous function $f(x)$ such that $f(f(x))$ is bounded.

pohoatza wrote: Does there exist a function $f: \mathbb{R}\rightarrow\mathbb{R}$ such that $f(x+f(y))=f(x)+\sin y$, for all reals $x,y$ ? $\sin{y}$ takes every value from $[-1;1]$ interval. So we can say that $f$ can get every value from $[f(x)-1;f(x)+1]$. After simple induction, we can say that $f$ is surjective. Then there exists some $x_0$ such that $f(x_0)=0$ . Put $x_0$ in the main equation : $f(x_0+f(y))=\sin{y}$ . This is right for any $y$ belongs to $\mathbb{R}$ . Because of $f$ is surjective and $x_0$ is fixed , we can say that $x_0+f(y)$ can get any value from $\mathbb{R}$ . Since $f$ is surjective and since last said fact, we can say that $f(x_0+f(y))$ takes every single value from $\mathbb{R}$, but on the other hand that expression is equal to $\sin{y}$ and can only take values from $[-1;1]$ . This is simple contradiction, so there doesn't exist such function. Too easy for P4

If $u\in f(\mathbb R)$ then by $P(u,\arcsin x)$ we have $[u-1,u+1]\subset f(\mathbb R)$, so by induction $f$ is surjective. Let $f(a)=2$, $f(b)=0$, and $f(c)=a-b$. $P(b,c)\Rightarrow\sin c=2$, a contradiction. So no such function exists.

Since $c + \epsilon\in \text{im}(f)$ whenever $c\in \text{im}(f)$ and $-1\le\epsilon\le 1$, by induction $f$ is surjective. Then $f^2$ is also surjective. But $P(0,y)\implies f(f(y)) = f(0) + \sin{y}$, a contradiction.